Question

A dishwasher has a mean life of 12 years with an estimated standard deviation of 1.25 years ("Appliance life expectancy," 2013). The life of a dishwasher is normally distributed. Suppose you are a manufacturer and you take a sample of 10 dishwashers that you made.
a.) State the random variable.


b.) Find the mean of the sample mean.


c.) Find the standard deviation of the sample mean.


d.) What is the shape of the sampling distribution of the sample mean? Why?


e.) Find the probability that the sample mean of the dishwashers is less than 6 years.


f.) If you found the sample mean life of the 10 dishwashers to be less than 6 years, would you think that you have a problem with the manufacturing process? Why or why not?

Answer 1

Answer:

a) Let X the random variable that represent the heights of a population, and for this case we know the distribution for X is given by:

$X \sim N(12,1.25)$  

Where $\mu=12$ and $\sigma=1.25$

b) $\mu_{\bar X}= 12$

c) $\sigma_{\bar X}= \frac{\sigma}{\sqrt{n}}= \frac{1.25}{\sqrt{10}}= 0.395$

d) Since the distribution for X is normal then we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu=12, \frac{\sigma}{\sqrt{n}}=0.395)$

e) $P(\bar X$

And we can use the z score given by:

$z =\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And using this formula we got:

$P(\bar X$

f) Yes is a problem since the probability for this event is very low so we can conclude that the event is not probable. And the sample mean is too low for the process.

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Part a

Let X the random variable that represent the heights of a population, and for this case we know the distribution for X is given by:

$X \sim N(12,1.25)$  

Where $\mu=12$ and $\sigma=1.25$

Part b

For this case the mean of the sample mean $\bar X$ is given by:

$\mu_{\bar X}= 12$

Part c

For this case the standard deviation for the sample mean would be given by:

$\sigma_{\bar X}= \frac{\sigma}{\sqrt{n}}= \frac{1.25}{\sqrt{10}}= 0.395$

Part d

Since the distribution for X is normal then we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu=12, \frac{\sigma}{\sqrt{n}}=0.395)$

Part e

For this case we want this probability:

$P(\bar X$

And we can use the z score given by:

$z =\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And using this formula we got:

$P(\bar X$

Part f

Yes is a problem since the probability for this event is very low so we can conclude that the event is not probable. And the sample mean is too low for the process.